Chemistry for the gifted and talented
Shapes of molecules and ions
Student worksheet: CDROM index 33SW
Discussion of answers: CDROM index 33DA
Topics
Valence shell electron pair repulsion (VSEPR) theory application.
Level
Very able post-16 students.
Prior knowledge
Valence shell electron pair repulsion (VSEPR) theory.
(An explanation of the VSEPR theory is given but this activity is best done after some
discussion of the theory in class.)
Resolving forces (questions 3 and 4).
Rationale
This activity draws some extra concepts and mathematical skills into the discussion of
molecular shape. A distinction between geometry around the central atom and the shape
of the molecule is made. Students are asked to apply the VSEPR theory to some
challenging examples, including species with unpaired electrons. Students are introduced
to the alternative model of hybridised atomic orbitals.
Use
This can be used as a follow up to a theory lesson on VSEPR theory for very able students.
It can also be used as a differentiated activity for the most able students in a mixed ability
group. If the students do not know how to resolve forces, they should leave out questions
3 and 4.
67
68
Chemistry for the gifted and talented
This symbol means those questions are best tackled as a discussion if a group of
students is doing this activity.
Student worksheet
Chemistry for the gifted and talented
Shapes of molecules and ions
An important theory used to explain and predict the shapes of molecules and ions is the Valence Shell
Electron Pair Repulsion (VSEPR) theory. This model is based on the idea that the pairs of electrons in
the valence, or outer, shell of the central atom repel each other. The species therefore adopts a
geometry that minimises the electron pair repulsion. The geometry around the central atom depends
on the effective number of electron pairs (double bonds count as one pair because both pairs in a
double bond are constrained to point in the same direction). The table below summarises the
relationship between the number of electron pairs and geometry.
Effective number of electron pairs
Geometry
2
Linear
3
Trigonal planar
4
Tetrahedral
5
Trigonal bipyramidal
6
Octahedral
7
Several –
Structure
including pentagonal
bipyramidal
If the electron pairs are not all the same, some distortion of the geometry can occur. Lone pairs are
thought to take up more space around the central atom than double bonds, which in turn take up
more space than single bonds. The amount of space taken up by a bonding pair may be affected by
the bond polarity.
continued on page 2
Student worksheet 33SW
Shapes of molecules and ions
Page 1 of 3
Chemistry for the gifted and talented
Questions
1. By drawing dot and cross diagrams and carefully counting the electrons or otherwise, suggest
the geometry and draw the shapes of the molecules and ions a)–e). Put your answers in a table
like the one below:
Molecule or ion
Effective number
Geometry around
of electron pairs in
central atom
Shape
valence shell of
central atom
a) SO2 (Hint: it is not linear)
b) O3
c) IF7
d) SF4 (There are five pairs of electrons around the sulfur. You need to think carefully about
the position of the lone pair.)
e) NO3–
2. Phosphorus(V) chloride in the solid state is thought to consist of the ions PCl4+ and PCl6–. Work
out the shape of these ions.
3. By considering each pair of electrons as an equal outward pull on the central nucleus and
resolving forces along one of the bonds, show that the bond angle in a symmetric trigonal
planar molecule, like BCl3, is 120°.
4. Using a similar method, show that the pure tetrahedral angle is 109.5°.
5. The HCH bond angle in ethene is 117.3°. Calculate how much greater the force of attraction
from the double bond is, compared to the single bonds. Comment on your answer.
continued on page 3
Student worksheet 33SW
Shapes of molecules and ions
Page 2 of 3
Chemistry for the gifted and talented
6. Phosphorus(V) chloride decomposes on heating to give PCl3 and Cl2. Explain whether
it is from the axial or equatorial positions that the chlorines break away.
7. Account for the bond angles in the following molecules:
Molecule
Bond angle
NH3
107°
NF3
102°
CH4
109.5°
CH3Cl
110.5° (HCH)
H2O
104.5°
H2S
92.2°
NO2
134°
An alternative model for explaining the shapes of molecules and ions uses hybrid atomic orbitals
(see further reading). The suggested hybridisation for different geometries is shown below.
Geometry
Linear
Trigonal
Tetrahedral
planar
Hybridisation
sp
sp2
Trigonal
Octahedral
bipyramidal
sp3
sp3d
sp3d2
Further reading
1. To read more about hybridised atomic orbitals, J. Keeler and P. Wothers, Why chemical reactions
happen, Oxford: Oxford University Press, 2003 (p82–93).
2. A good website to visit is:
www.psigate.ac.uk/newsite/reference/plambeck/chem1/p02201.htm (accessed June 2007).
Student worksheet 33SW
Shapes of molecules and ions
Page 3 of 3
Discussion of answers
Chemistry for the gifted and talented
Shapes of molecules and ions
1. State the geometry and draw the shapes of the following molecules and ions.
Question
Molecule
or ion
Effective
number of
electron pairs
Geometry
around central
atom
a)
SO2
3 (sulfur can
‘expand’ its octet)
Trigonal planar
b)
O3
3 (double bonds
count as 1 pair)
Trigonal planar
c)
IF7
7
Pentagonal
bipyramidal
d)
SF4
5
Trigonal
bipyramidal
(distorted)
e)
NO3–
3
Trigonal planar
Shape
Note: In SF4 the lone pair could potentially be in an axial position or an equatorial position. The
configuration with the lone pair in the equatorial position allows more space for the bulky lone pair,
as there are only two other pairs at 90° to it.
Axial lone pair
Equatorial lone pair
The lone pair takes up more room than bonding pairs of electrons, so the shape is slightly distorted
with the bonding electrons pushed away from the lone pair.
The FSF bond angle (for axial fluorines) is 173.1°.
continued on page 2
Discussion of answers 33DA
Shapes of molecules and ions
Page 1 of 4
Chemistry for the gifted and talented
SF4, SO2 and IF7 are all examples of where the central atom has ‘expanded its octet’ – ie contains
more than eight electrons in its valence shell. This is easier in the third period of the Periodic Table
than in the first or second periods because of the availability of d atomic orbitals..
The nitrate ion is thought to have a conjugated structure (as shown in the second structure), in a
similar way to carboxylate ions.
2. Phosphorus(V) chloride in the solid state is thought to consist of the ions PCl4+ and PCl6–. Work
out the shape of these ions.
Tetrahedral
and octahedral
3. By considering each pair of electrons as an outward pull on the central nucleus and resolving
forces along one of the bonds, show that the bond angle in a symmetrical trigonal planar
molecule like BCl3 is 120°.
Each arrow represents a force of y. Resolving forces in the horizontal direction, 2ycosθ = y because
the forces on the nucleus must be balanced, cosθ =
1
2
therefore θ = 60°.
180°–60° = 120°, or bond angle = 2θ = 120°.
4. Using a similar method, show that the pure tetrahedral angle is 109.5°.
The method is the same, but this time there are three arrows pointing to the left so 3ycosθ = y
cosθ = 1/3 therefore θ = 70.5°
Thus bond angles 180°–70.5° = 109.5°.
continued on page 3
Discussion of answers 33DA
Shapes of molecules and ions
Page 2 of 4
Chemistry for the gifted and talented
5. The HCH bond angle in ethene is 117.3°. Calculate how much greater the force of attraction
from the double bond is, compared to the single bonds. Comment on your answer.
Let d be the pull from the double bond and let s be the pull from the single bond. Resolving forces
along the double bond gives:
d=2scos58.65°
d= 1.04s.
This is surprisingly small, given that there are four electrons in the double bond compared with only
two in the single bond. The explanation for the surprisingly small attraction along the axis of the
double bond must be that, because the electron density of the π electrons is largely above and below
the plane of the molecule, most of the attraction of the π electrons is perpendicular to the attraction
of the σ electrons.
6. Phosphorus(V) chloride decomposes on heating to give PCl3 and Cl2. Explain whether it is from
the axial or equatorial positions that the chlorines break away.
Chlorines break away from the axial positions since these are more crowded, with three neighbours
at 90° rather then two neighbours at 90° and two at 120° for the equatorial chlorines.
7. Account for the bond angles in the following molecules:
Molecule
Bond angle
NH3
107°
NF3
102°
CH4
109.5°
CH3Cl
110.5° (HCH)
H2O
104.5°
H2S
92.2°
NO2
134°
The NH3 bond angle is 107°. This is distorted from the pure tetrahedral angle by the lone pair taking
up more space than the single bonds.
The NF3 bond angle is 102°. There is more distortion than for NH3 because the single bonds are
taking up less room, close to the nitrogen. Fluorine is more electronegative than hydrogen and the
electron density in the N–F bond is skewed towards the fluorine.
The CH4 bond angle is 109.5°. This is the undistorted tetrahedral angle.
continued on page 4
Discussion of answers 33DA
Shapes of molecules and ions
Page 3 of 4
Chemistry for the gifted and talented
The (HCH) bond angle in CH3Cl is 110.5°. The HCH bond angle is wider than in CH4 which suggests
that the C–Cl single bond takes up less space close to the carbon. This is due to the C–Cl bonds
polarity with δ– on the chlorine, chlorine being more electronegative than carbon (and hydrogen). The
C–H bond has more electron density around the carbon because the carbon is δ–, carbon being more
electronegative than hydrogen.
The H2O bond angle is 104.5°. There is greater distortion than in NH3 because there are two lone
pairs on the oxygen.
The H2S bond angle is 92.2°. There is a surprisingly large difference between this and the bond angle
in H2O. Sulfur is less electronegative than oxygen, so the S–H bond takes up less space around the
central atom than the O–H bond. In terms of hybridisation, the reduced bond angle in H2S implies
the bonds that are formed are much closer to p atomic orbitals, rather than sp3 hybrid atomic
orbitals. A reason for this might be weaker bonds, therefore less return on the investment in energy
to hybridise the atomic orbitals.
The NO2 bond angle is 134°. This is a perhaps a surprising value. The first step towards explaining
this is to propose a structure for NO2 that is consistent with this bond angle.
Around the nitrogen there is a double bond, a single bond and an unpaired electron (remember that
NO2 dimerises to form N2O4). The geometry is trigonal planar but the unpaired electron repels the
bonding pairs less than a pair of electrons so the bond angle is greater than 120°.
For more discussion on NO2 visit:
http://dbhs.wvusd.k12.ca.us/webdocs/VSEPR/Odd-Electron-Molecules.html
(accessed May 2007).
Discussion of answers 33DA
Shapes of molecules and ions
Page 4 of 4